year 12 advanced mathematics sample...

of 15 Alternative Education Equivalency Assessments (AEEA)

Year 12 Advanced Mathematics Sample Questions

Year 12 Advanced Mathematics

Sample Questions Alternative Education Equivalency Assessments (AEEA)

ADVANCED MATHEMATICS

The following examples show the types of items in the test, but do not necessarily indicate the full

range of items or test difficulty. For the Advanced Mathematics test, you may use a silent,

battery-operated, non-programmable scientific calculator (not CAS/graphics calculator) and a

ruler. See the Solutions pages for answers to these sample questions.

Formulae

The following formulae may be used in your calculations:

Quadratic Equations

If ax2 bx c 0 then x

b (b2 4ac)

2a

Series

where a is the first term, L is the last, d is the common difference and r is the common ratio

Arithmetic

a (ad) (a2d) ... (a (n1)d)n

2(2a (n1)d)

n

2(a L)

Geometric

aar ar2 ...arn1 a(1 rn)

1 r, r 1

Space & Measurement

In any triangle ABC,

1sin

2Area ab C

Trapezium: Area = height, where a and b are the lengths of the parallel sides

Prism: Volume = Area of base height

Cylinder: Total surface area = Volume =

Pyramid: Volume = area of base height

sin sin sin

a b c

A B C

2 2 2 2 cos a b c bc A

2 2 2

cos2

b c aA

bc

1( )

2a b

22 2r h r 2r h

1

3



Space & Measurement (cont’d)

Cone: Total surface area = , s is the slant height Volume =

Sphere: Total surface area = Volume =

Volume of solids of revolution about the axes: and

Rate: If y’ = ky, then y = Ae kx

Temperature conversion formula

Degrees Celsius to degrees Fahrenheit: ( 1.8) 32F C

Theorem of Pythagoras

In any right-angled triangle: 2 2 2c a b

Index laws

For and m ,n real,

For m an integer and n a positive integer

Calculus

Function notation Leibniz Notation

Product

rule

Quotient

rule

Chain rule

Fundamental Theorem of Calculus: and

Standard Derivatives

If y f (x) xn , then y '

dy

dx f '(x) nxn1

If y f (x) ex , then

dy

dx f '(x) ex

If y f (x) log

ex then y '

dy

dx f '(x)

1

x

If y f (x) sin(ax), then y '

dy

dx f '(x) a cos(ax)

If y f (x) cos(ax), then y '

dy

dx f '(x) a sin(ax)

2r s r 21

3r h

24 r 34

3r

2y dx2x dy

, 0a b

m n m na a a ( )m m ma b ab ( )m n mna a

1m

ma

a

m

m n

n

aa

a

0 1a

m

mn nmna a a

y y y y

( ) ( )f x g x ( ) ( ) ( ) ( )f x g x f x g x uvdu dv

v udx dx

( )

( )

f x

g x 2

( ) ( ) ( ) ( )

( ( ))

f x g x f x g x

g x

u

v 2

du dvv u

dx dx

v

( ( ))f g x ( ( )) ( )f g x g x ( ) ( )andy f u u g x dy du

du dx

( ) ( )x

a

df t dt f x

dx ( ) ( ) ( )

b

af x dx f b f a



Standard Integrals

= , , and if

= = ,

= , = ,

Probability laws

Trigonometry

In any right-angled triangle:

sin θ = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

cos θ = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

tan θ = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒

𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

Growth, decay and interest formulae

Simple growth or decay: (1 )A P ni

Compound growth or decay: (1 )nA P i

Where:

A = amount at the end of n years

P = principal

n = number of years

r % = interest rate per year, i = r

100

dxxn

1

1

1

nxn

1n 0x 0n

dxx

1,ln x 0x dxeax

axe

a

10a

axcos dx axa

sin1

0a axsin dx axa

cos1

0a

( ) ( ) 1P A P A

( ) ( ) ( ) ( )P A B P A P B P A B

( ) ( ) ( / ) ( ) ( / )P A B P A P B A P B P A B

)Pr(

)Pr()/Pr(

B

BABA

Opposite side

Adjacent side

Hypotenuse



Growth, decay and interest formulae (cont’d)

Compound interest, where the interest is compounded t times per year:

nt

t

iPA )1(

Where:

t number of interest periods per year

Future value of an annuity: [(1 ) 1]nx i

Fi

contributions at end of each period

OR [(1 ) 1] (1 )nx i i

Fi

contributions at beginning of each period

Where:

F = future value of annuity

i = interest rate per compounding period, as a decimal fraction

n = number of compounding periods



Real functions

Example 1 For the basic following functions: f(x) = x

x

1

12 and h(x) = 1 – 2x find the

composite function , ( ( ))f h x in simplest terms: (2 marks)

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Linear functions

Example 2 The line 2y + x = 4 is reflected across the x axis. Sketch the original line and its

reflection (clearly marking coordinates of any intercepts) then find the equation of

the reflected line. (3 marks)

The quadratic polynomial and the parabola

Example 3 Find the coordinates of the turning point of the parabola y = x2 – 4x – 5 (2marks)

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Plane geometry – geometrical properties

Example 4 An equilateral triangle is inscribed in a circle of radius 3cm. Calculate the unshaded

area as shown below (correct to 2 decimal places) (3 marks)

Tangent to a curve and derivative of a function

Example 5 Find the gradient of the curve f(x) = 2e3x at the point where x = 1

(correct to 2 decimal places) (2 marks)

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Coordinate methods in geometry

Example 6 The vertices of ∆ABC are A(1,2), B(6,-1) and C(2,-2). Use your knowledge of the

properties of a right angled triangle to show that ∆ABC is a right angled triangle.

(2 marks)

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Applications of geometrical properties

Example 7 Given AB = 5 units, ED = 3 units and AD = 4 units, find the length of DC in the

diagram below. (2 marks)

(diagram not drawn to scale)

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Geometrical applications of differentiation

Example 8 Find the equation of the normal to the curve 2)2( xy at the point where 3x .

(4 marks)

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Integration

Example 9 Find the exact value of

2

0

2)12(

1

xdx (2 marks)

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Trigonometric functions (including applications of trigonometric ratios)

Example 10 The function f(Ѳ) = 1 + sin2Ѳ is defined for Ѳ є [0,2 ]. Write down the maximum

value of f(Ѳ) and the values of Ѳ for which it occurs. (3 marks)

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Logarithmic and exponential functions

Example 11 An insect population grows according to the rule P = 2loge(t + 2) where P is the

population, in millions, t years after the population was first estimated.

According to this rule:

a) What was the population when first estimated? (1 marks)

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b) How long will it take for the population to reach 5 million? (correct to 2 decimal

places) (2 marks)

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Applications of calculus to the physical world

Example 12 With wind assistance, a balloon ascends at an acceleration of 2t m sec-2 (where t is

the time in seconds after release). If the balloon is stationary until it is released

from a height of 1 metre above ground level, how long will it take to reach a height

of 100 metres? (correct to 2 decimal places) (4 marks)

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Probability

Example 13 A tennis player wins 80% of her matches. To the nearest %, what is the probability

she will win at least 4 of her next 5 matches? (2 marks)

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Series and series applications

Example 14 A “not so wise” boss agreed to pay a worker $1 on the 1st day, $2 on the 2nd day, $4

on the 3rd day and so on.

a) Show that this form of payment is a geometric sequence. (1 marks)

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b) How many dollars would the boss have to pay on the 20th day? (2 marks)

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Year 12 Advanced Mathematics Sample Questions Solutions

Example 1 solution

2(1 2 ) 1 1 4( ( )) (1 2 )

1 (1 2 ) 2 2

x xf h x F x

x x

Example 2 solution

New equation : 2y – x = -4 or x – 2y = 4 or –x + 2y = -4

Both intercepts (or 2 points) must be shown on graphs

Example 3 solution

y = (x – 2)2 – 5 – 4 OR y = (x –5)(x + 1)

y = (x – 2)2 – 9 TP at x = 2

15 = 2, y = (2 – 5)(2 + 1) = -9

TP (2, -9) TP (2, -9)



Example 4 solution

Area of 3 triangles side length 3cm, α = 120o = 3 x ½ (3)(3)sin120o = 11.6913

Area of circle 𝐴 = 𝜋𝑟2 = x 9 = 28.2743

Unshaded area = 28-2743 – 11.6913 = 16.58cm2

Example 5 solution

f’(x) = 6e3x

f’(1) = 6e3 = 120.51

Example 6 solution

m(AC) = 12

22

= -4, m(BC) =

4

12

= ¼

-4 x ¼ = -1 so sides AC and BC are at right angles ∆ABC is a right angled triangle

OR

AC 2 = (2-1)2 + (-2-2)2 = 17 AB2 = (6-1)2 + (-1-2)2 = 34 BC2 = (2-6)2 + (-2- -1)2 = 17

BC2 + AC2 = AB2 so ∆ABC must be a right angled triangle



Example 7 solution

Let DC = x

3

5 =

x

x4

5x = 12 + 3x

2x = 12 x = 6

Example 8 solution

dx

dy= 2(x -2)

Gradient of tangent is 2(3 – 2) = 2 at x = 3

Gradient of the normal is:

,2

1

3

1,

2

1

x

ym

Equation of the normal 52 xy

Example 9 solution

2

0

2)12(

1

xdx = -½ (2x + 1)-1 2

0

= -( 101

) - (- 21

)

= 52

Example 10 solution

Max value of 2

when:

2Ѳ = 2

,

2

+ 2

Ѳ = 4

,

4

+

Ѳ = 4

,

4

5 (no working or sketch required - marks only for answers)



Example 11 solution

a) 1.39 million

b) 5 = 2loge(t + 2)

e2.5 = t + 2

t = 10.18 years

Example 12 solution

Let x = height above ground level

a = ..

x = 2t

v = .

x = t2 dt = t2 + c

At t = 0, v = 0 c = 0

.

x = t2

x = t2 dt =

3

1t3 + c1

At t = 0, x = 1 c1 = 1

x = 3

1t3 + 1

At x = 100 = 3

1t3 + 1

t = 3 297 = 6.67 seconds

Example 13 solution

4

5(0.8)4(0.2)1 + (0.8)5 (1 or both terms correct)

= 0.4096 + 0.3277 = 74%

Example 14 solution

a) Common ratio = 2 i.e. 1

2 =

2

4

b) a = 1, r = 2, t20 = ar19 =1 x 219

= $524,288



Notes

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